Consider square with vertices at (1,1), (-1,1), (-1, -1), and (1, -1).
S is the region which consists of all points inside the square which are nearer to the origin than any edge of this square.
Determine the area of S.
Again consider 1/8 of the square, a 1-1-√2 triangle with vertices:
(0,0), (0,1), (1,1).
Consider a point (r,Θ). It is distance r from the center,
and 1 - r*cos(Θ) from the closest edge.
r = 1 - r*cos(Θ)
r = 1/(1 + cos(Θ))
The area inside the locus of points that forms the boundary is
integral from 0 to pi/4 of (1/2)*r^2 * dΘ
or integral of (1/2)(1/(1 + cos(Θ)))^2 dΘ
Wolfram Alpha says this is:
integral (1/2)(1/(1 + cos(Θ)))^2 dΘ
= (sin(Θ)*(cos(Θ) + 2))/(6*(cos(Θ) + 1)^2) + constant
or for the definite integral with the limits in question:
= 1/6 (4 sqrt(2) - 5) ≈ 0.10948
And multiplying that result by 8 gives the same answer as found previously:
≈ 0.87584
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Posted by Larry
on 2024-03-01 11:34:42 |
Polar coordinates
